Expand binomially to prove trigonometric identity

Question

Prompt: By expanding $\left(z+\frac{1}{z}\right)^4$ show that $\cos^4\theta = \frac{1}{8}(\cos4\theta + 4\cos2\theta + 3).$

I did the expansion using binomial equation as follows

$$\begin{align*} \left(z+\frac{1}{z}\right)^4 &= z^4 + \binom{4}{1}z^3.\frac{1}{z} + \binom{4}{2}z^2.\frac{1}{z^2} + \binom{4}{3}z^3.\frac{1}{z}+\frac{1}{z^4}\\ &=z^4+4z^2+6+\frac{4}{z^2}+\frac{1}{z^4}\\ &=z^4+\frac{1}{z^4}+4\left(z^2+\frac{1}{z^2}\right) + 6. -(eqn 1) \end{align*} $$

I'm not sure how to go on about rest of the problem.

[update] Reading comments, I tried assuming $z = e^{i\theta}$

$2\cos\theta = e^{i\theta} + e^{-i\theta}$

$(2\cos\theta)^4 = (e^{i\theta} + e^{-i\theta})^4$

$=e^{4i\theta} + e^{-4i\theta} + 8(e^{2i\theta}+e^{-2i\theta})+6$ (from eqn 1)

Answer 1

HINT:

Set $z=e^{i\theta}$ and use Euler's formula.

Answer 2

hint

Take now $z=e^{i\theta} $

and use Euler identity

$$\cos (x)=\frac {e^{ix}+\frac {1}{e^{ix}}}{2} $$

Answer 3

Let $\cos \theta=z+\dfrac1z.$
Then $\cos2\theta=2\cos^2\theta-1=2z^2+3+\dfrac2{z^2}$ and similarly you can find $\cos 4\theta.$
Compare them with your binomial identity.
Good luck.

Answer 4

Following your idea, you can suppose that $|z|=1$ and then $z=\text{cis}\theta$ and $z^{-1}=\text{cis}(-\theta)$

$$z+z^{-1}=2\cos\theta \to (z+z^{-1})^4=2^4\cos^4\theta$$

on the other hand:

$$z^2+z^{-2}=(z+z^{-1})^2-2=2^2\cos^2\theta-2=2\cos2\theta$$

$$z^4+z^{-4}=(z^2+z^{-2})^2-2=(2\cos2\theta)^2-2=2\cos4\theta$$

Replacing on your last equation you get:

$$2^4\cos^4\theta=2\cos4\theta+8\cos2\theta+6\\ \cos^4\theta=\frac{1}{8}(\cos4\theta+4\cos2\theta+3)$$